Concepts of Physics

IIT JEE Physics (1978-2016: 39 Years) Topic-wise Complete Solutions

Dynamics of Rigid Bodies with Fixed Axis of Rotation

Interesting Questions

Question: If a raw egg and a boiled egg are spinned together with same angular velocity on the horizontal surface then which one will stops first? (Source FB)

Answer: Consider the rotation of hard boiled egg. The friction is the only reason which can stop it. The time to stop will depend on initial angular velocity and torque due to frictional force. Now, consider the raw egg. It is not a rigid body because fluid start rotating relative to the shell. It stops rotating because of (i) torque due to frictional force and (ii) loss of energy due to viscous drag. Going by this logic, raw egg should stop first if frictional forces are equal in two cases. It will be interesting to do the real experiment and get the result? We expect something strange when boiled egg is rotated very fast!

Problems from IIT JEE

Problem (IIT JEE 2014): Dynamics of Rigid Bodies A horizontal circular platform of radius 0.5 m and mass 0.45 kg is free to rotate about its axis. Two massless spring toy-guns, each carrying a steel ball of mass 0.05 kg are attached to the platform at a distance 0.25 m from the centre on its either sides along its diameter (see figure). Each gun simultaneously fires the balls horizontally and perpendicular to the diameter in opposite directions. After leaving the platform, the balls have horizontal speed of 9 m/s with respect to the ground. The rotational speed of the platform in rad/s after the balls leave the platform is,

Solution: Consider the balls and the platform as a system. There is no external torque on the system about its centre. Hence, angular momentum of the system about its centre is conserved. Initial and final angular momentum of the system are, \begin{align} &L_i=0,&&L_f=mvr+mvr+I\omega=2mvr+\tfrac{1}{2}MR^2\omega.\nonumber \end{align} The conservation of angular momentum, $L_i=L_f$, gives, \begin{align} \omega=-\frac{4mvr}{MR^2}=-\frac{4(0.05)(9)(0.25)}{0.45(0.5)^2}={-4}\;\mathrm{rad/s}.\nonumber \end{align}